Unit 6 Stoichiometry - Central Dauphin School District

percent yield 2013 answers.notebook

February 28, 2013

Unit 6 Stoichiometry Percent Yield

Steps for finding PERCENT YIELD: 1. Balance the chemical equation 2. Find the limiting reactant 3.Find the theoretical yield 4.Find the actual yield 5.Find the percent yield

1

percent yield 2013 answers.notebook

February 28, 2013

% yield = mass Actual Yield

X 100%

mass Theoretical Yield

*The actual yield is often given in the problem...is occasionally referred to

as the experimental yield.

*The theoretical yield is determined by solving a problem using the

stoichiometry map, usually a 3step

gramgram problem.

maximum amount of product that could possibly form

PERCENT YIELD

Theoretical yield

Used to help you figure out where to put the compounds in a 3step stoich problem

L 7

needed to do a stoichiometry problem

amount of product that forms

experimentally in the lab

actual yield

reactant used to determine the theoretical yield

Limiting Reactant

Balanced Equation

this step uses coefficients from balanced equation

mole ratio

2

percent yield 2013 answers.notebook

February 28, 2013

Nitrogen dioxide, NO2 , can be converted to dinitrogen pentoxide, N2O5 , by reacting it with ozone, O3 .

2NO2(g) + O3(g)

N2O5(s) + O2(g)

Calculate the percent yield for a reaction in which 0.38 g of NO2 reacts with excess O3 and 0.36 g of N2O5 is recovered.

theoretical yield = actual yield =

% yield = mass Actual Yield X 100%

mass Theoretical Yield

theoretical yield = 0.45g actual yield = 0.36 g

% yield =

X 100%

3

percent yield 2013 answers.notebook

February 28, 2013

packet:

1. Determine the percent yield for the reaction between 3.74 g Na and excess O2 if 5.34 g of Na2O2 is recovered.

balanced equation:

% yield =

X 100%

Mass in grams

L

(a) Mass of watch glass

61.19 g

A

(b) Mass of watch glass + potassium iodide

62.01g

B

(c) Mass of potassium iodide b a

0.82 g

(d) Mass of watch glass

61.19 g

D

(e) Mass of watch glass + lead nitrate

62.38 g

A

(f) Mass of lead nitrate e a

1.19 g

T

(g) Mass of filter paper

(h) Mass of filter paper and precipitate PbI2 after dried*

1.09 g

A

2.10 g

(i) Mass of precipitate after dried* also referred to as the experimental yield h g

1.01 g

4

percent yield 2013 answers.notebook

February 28, 2013

2. In this reaction, one of the reagents was a limiting reagent. Using your knowledge of chemical equations and limiting reagents, determine which was the limiting reactant for this experiment.

Use (c) from the data table and (f) from the data table to do a 3step LR problem.

0.82 g KI

PbI2

PbI2 PbI2

1.19 g Pb(NO3)2

PbI2

PbI2

PbI2

3. Using the LR from #2, write down the theoretical yield of PbI2

% yield = mass Actual Yield

X 100%

mass Theoretical Yield

5

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download